Given an arbitrary set $A$ of real numbers. I want to decide if the set $A$ is infinite.
My question is: What is the logical reason to use a proof by contradiction.
I can think for example that there is no known method to prove this directly. But I am not convinced with this reason.
There's a subtle distinction here, in that you can prove a negation without using contradiction.
Proof-by-contradiction is what we usually call the use of the double-negation law: "from $\neg \neg P$, I can deduce $P$".
But you're trying to deduce something of the form $\neg P$ ("the set $A$ is not finite"), and there's one obvious way to do that without using double-negation: suppose $P$ and then deduce a contradiction. That is automatically a proof of $\neg P$, without ever having to invoke the axiom $\neg \neg P \to P$.
If you're familiar with the usual definition of $\neg P$ as $P \to \bot$ (where $\bot$ is the "false" symbol), then this becomes clearer. How else would you try and prove $P \to \bot$, without assuming $P$ and showing $\bot$?
So my answer to your question is: you're trying to prove the statement that "if $A$ is finite then I can prove false", and the easiest way to start doing this is to assume $A$ is finite and then just directly prove false.
It is a proof technique, based on the rule of Indirect proof (aka : Proof by contradiction).
The gist is : the rules of logic are sound, i.e. they allow us to derive only true conclusions from true premises.
Thus, if we assume $\lnot A$, i.e. that statement $A$ is False, and we derive by way of correct logical inferences, a contradiction, i.e. a statement that is always False, we have only two possibilities :
either (i) we have made a logical mistake, or
(ii) our original assumption is wrong.
Thus, if we agree on the correctness of the logic, we are forced to conclude with (ii) :
the assumption that statement $A$ is False is wrong, i.e. $A$ must be True.
To be more specific with your concern (for what I've understood) : dealing with "the infinite" is not easy.
How can we show that a certain set is infinite ? In this case a very common proof technique is that of Proof by contradiction: assume that the said set is finite and derive a contradiction with the definition (or previously known properties) of the set.
In practice, it is easier to prove the existence of something than to prove its non-existence.
If you have the list of all primes up to $n$, you can prove constructively that there is a larger prime. The proof will rest on the list.
It is harder to prove out of the blue that "there is no largest prime", because you have nothing to start from. You cannot even speak of that largest prime, as you are precisely proving that there is none.
But the two approaches are only superficially different, as "there is no largest prime" is equivalent to "for all primes, there is a larger prime".
$$\nexists p\in\mathbb P:\forall q\in\mathbb P:p\ge q\iff \forall p\in\mathbb P:\lnot(\forall q\in\mathbb P:p\ge q)\iff \forall p\in\mathbb P:\exists q\in\mathbb P:p<q.$$
Let me start it by giving you the definition of " proof by contradiction " :
When we are supposed to prove something which clearly has a specific alternative answer. And also you already know the answer and you are supposed to prove it no matter however obvious it may seem.
Like in your example: Either your set $A$ can be finite or infinite. So, where one has binary possibilities of answer, one can think of prooving it by contradiction and it has been said as one of the finest weapons of mathematicians by G.H. Hardy saying "It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game."
Even basic principle behind proof by contradiction can be projected in "Sherlock Holmes" in the novel The Sign of the Four (1890) by Sir Arthur Conan Doyle which says "When you have eliminated the impossible, whatever remains, however improbable, must be the truth"
The reason most people do proof by contradiction isn't really logical.
When you do a direct proof, you have $n$ assumptions and $1$ predetermined statement to prove.
When you do proof by contradiction, you have $n+1$ assumptions and you succeed when you prove any false or contradictory statement.
Some people find the second scenario more freeing. If you are interested in the subject you are proving about, it is worthwhile to unwind your proof by contradiction to see what a direct proof would look like.
I would say the 'logical', or at 'least conceptually 'intuitive' reason for using a proof by contradiction is because you are trying to prove that something is not the case, i.e. some kind of negation.
Now, in this case, we can naturally think of 'infinite' as 'not finite'
Of course, there are more 'positive' formulations of infinite as well ... and so why would 'infinite' be the 'negative' one, and 'finite' be the 'positive' one? Still, I think the more natural way to think about 'infinity' is in a 'negative' way.
Moreover, we can do a lot more with something being finite. Being finite, allows us to say: ok, so it is some number ... and we're off to the races trying to see what follow from that.... hopefully a contradiction!
Suppose an hypothesis, say H, implies a contradiction, say (A&~A), which means that the conditional : H ---> (A&~A) is true.
Is it possible H to be true?
If H is true, the antecedent of the conditional ( namely H itself) is true and the consequent of the conditional ( namely : A & ~A) is false.
But such a conditional is false by definition ( as one can see when looking at the defining truth table of the " if ..then" operator).
So, H must be false.
Another justification is provided by the fact that the conditional
[ H --> (A & ~A)] --> ~H
is a tautology, which justifies the validity of the following reasoning :
H implies a contradiction, therefore H is false.
A third reason is that, if you allow the hypothesis H to remain in your theory , even though it implies a contradiction, your theory will contain all possible propositions whithout exception. Indeed, H implies a contradition and a contradiction implies anything ( as one can verify using a truth table). And since implication is a transitive relation, H will ( indirectly) imply anything, even its own negation ( ~H).
